Hi,
sorry if my description was too short / unclear.
> I have a list of 7 data frames, each data frame having 24 rows (hour of
> the day) and 5 columns (weeks) with a total of 5 x 24 values
[1]
week1 week2 week3 ...
1 x a m ...
2 y b n
3 z c o
. . . .
. . . .
. . . .
24 . . .
[2]
week1 week2 week3 ...
1 x2 a2 m2 ...
2 y2 b2 n2
3 z2 c2 o2
. . . .
. . . .
. . . .
24 . . .
[3]
...
.
.
.
[7]
...
I now would like to extract e.g. all week2 columns of all data frames in
the list and combine them in a new data frame using cbind.
new data frame
week2 ([1]) week2 ([2]) week2 ([3]) ...
a a2 .
b b2 .
c c2 .
.
.
.
I will then do further row-wise calculations using e.g. apply(x,1,mean),
the result being a vector of 24 values.
I have not found a way to extract specific columns of the data frames in
a list.
As mentioned I can use
sapply(list_of_dataframes,"[",1:24)
which will pick the first 24 values (first column) of each data frame in
the list and arrange them as an array of 24 rows and 7 columns (7 data
frames are in the list).
To pick the second column (week2) using sapply I have to use the next 24
values from 25 to 48:
sapply(list_of_dataframes,"[",25:48)
It seems that sapply treats the data frames in the list as vectors. I
can of course extract all consecutive weeks using consecutive blocks of
24 values, but this seems cumbersome.
The question remains, how to select specific columns from data frames in
a list, e.g. all columns 3 of all data frames in the list.
Reformatting (unlist(), dim()) in one data frame with one column for
each week does not help, since I'm not calculating colMeans etc, but
row-wise calculations using apply(x,1,FUN) ("applying a function to
margins of an array or matrix").
Thanks for you help and suggestions!
Wolfgang
On 08/02/16 18:00, Dénes Tóth wrote:
> Hi,
>
> Although you did not provide any reproducible example, it seems you
> store the same type of values in your data.frames. If this is true, it
> is much more efficient to store your data in an array:
>
> mylist <- list(a = data.frame(week1 = rnorm(24), week2 = rnorm(24)),
> b = data.frame(week1 = rnorm(24), week2 = rnorm(24)))
>
> myarray <- unlist(mylist, use.names = FALSE)
> dim(myarray) <- c(nrow(mylist$a), ncol(mylist$a), length(mylist))
> dimnames(myarray) <- list(hour = rownames(mylist$a),
> week = colnames(mylist$a),
> other = names(mylist))
> # now you can do:
> mean(myarray[, "week1", "a"])
>
> # or:
> colMeans(myarray)
>
>
> Cheers,
> Denes
>
>
> On 02/08/2016 02:33 PM, Wolfgang Waser wrote:
>> Hello,
>>
>> I have a list of 7 data frames, each data frame having 24 rows (hour of
>> the day) and 5 columns (weeks) with a total of 5 x 24 values
>>
>> I would like to combine all 7 columns of week 1 (and 2 ...) in a
>> separate data frame for hourly calculations, e.g.
>>> apply(new.data.frame,1,mean)
>>
>> In some way sapply (lapply) works, but I cannot directly select columns
>> of the original data frames in the list. As a workaround I have to
>> select a range of values:
>>
>>> sapply(list_of_dataframes,"[",1:24)
>>
>> Values 1:24 give the first column, 25:48 the second and so on.
>>
>> Is there an easier / more direct way to select for specific columns
>> instead of selecting a range of values, avoiding loops?
>>
>>
>> Cheers,
>>
>> Wolfgang
>>
>> ______________________________________________
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
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